Problem: The equation of a circle $C$ is $x^2+y^2+12x-18y+92 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2+12x) + (y^2-18y) = -92$ $(x^2+12x+36) + (y^2-18y+81) = -92 + 36 + 81$ $(x+6)^{2} + (y-9)^{2} = 25 = 5^2$ Thus, $(h, k) = (-6, 9)$ and $r = 5$.